Lecture 05 Analysis (I)
Time Response and State Transition Matrix
5.1 State Transition Matrix
5.2 Modal decomposition --Diagonalization
5.3 Cayley-Hamilton Theorem
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d
x(t )  Ax(t )  Bu (t )
dt
y(t )  Cx(t )  Du (t )
The behavior of x(t) et y(t) :
1. Homogeneous solution of x(t)
2. Non-homogeneous solution of x(t)
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Homogeneous solution
x (t )  Ax(t )
1
sX ( s)  x(0)  AX ( s)
1
x(t )  L [( sI  A) ]x(0)
 e At x(0)
X ( s)  ( sI  A) 1 x(0)
State transition matrix

(t )  e At  L1[( sI  A) 1 ]
x(t0 )  e At0 x(0)
x(0)  e
 At0
x(t0 )
At  At0
x(t )  e e
x(t0 )  e
A ( t t 0 )
x(t0 )  (t  t0 ) x(t0 )
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
Properties
1.
1
1
(t )  e  L [( sI  A) ]
At
(0)  I
1
2.
 (t )  (t )
3.
x(0)  (t ) x(t )
4.
(t 2  t1 ) (t1  t0 )  (t 2  t0 )
5.
(t )  (kt)
k
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Non-homogeneous solution
d
x(t )  Ax(t )  Bu (t )
dt
y(t )  Cx(t )  Du (t )
sX ( s )  x(0)  AX ( s )  BU ( s )
( sI  A) X ( s )  x(0)  BU ( s )
1
1
X ( s )  ( sI  A) x(0)  ( sI  A) BU ( s )
x(t )  L1[( sI  A) 1 ]x(0)  L1[( sI  A) 1 BU ( s )]
t
x(t )   (t ) x(0)    (t   ) Bu ( )d
0
Convolution
Homogeneous
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t
x(t )  (t ) x(0)   (t   ) Bu ( )d
0
t
x(t )  (t  t0 ) x(t0 )   (t   ) Bu ( )d
t0
t
y (t )  C(t  t0 ) x(t0 )   C(t   ) Bu ( )d  Du (t )
t0
Zero-input response
Zero-state response
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Example 1
1   x1  0
 x1   0
 x    2  3  x   1u (t )
 2   
 2 
let
x(0)  0 0
T
t
2t

2
e

e
(t )  L1[( sI  A) 1 ]  e At  
t
 2t

2
e

2
e

e 1  e 2t 

 e t  2e 2t 
t
x(t )  (t ) x(0)   (t   ) Bu ( )d
0
Ans:
1
3
 x1    2e t  e  2t 
2
x    2

t
 2    2e  2e 2t 
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L1[( sI  A) 1 BU (s)]
7
1   x1  0
 x1   0
 x    2  3  x   1u (t )
 2   
 2 
let
x2 (0)
s
x1 (0)
s
x(0)  0 0
T
1
u
s 1
x2
s 1
x1 1
y
3
Using Maison’s gain formula
2
  1  3s 1  2 s 2
s 1 (1  3s 1 )
s 2
s 2
x1 ( s ) 
x1 (0) 
x 2 ( 0) 
U (s)



 2s 2
s 1
s 1
x2 ( s ) 
x1 (0) 
x 2 ( 0) 
U (s)



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How to find State transition matrix

1
1
(t )  e  L [( sI  A) ]
Methode 1:
Methode 2:
At
(t )  L1[( sI  A) 1 ]
(t )  e At
Methode 3: Cayley-Hamilton Theorem
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Methode 1:
1
1
(t )  L [( sI  A) ]
1
0   x1  0 0
 x1   0
 x    0  4 3   x   1 0   u1 
 2 
 2  
 u 
 x3   1  1  2  x3   0 1  2 
 x1 
 y1 (t )  1 0 0  
 y (t )  0 0 1  x2 
x 
 2  
 3
( sI  A) 1 
adj ( sI  A)
sI  A
 s 2  6 s  11 s  2
3 
1


2


3
s

2
3
s

s ( s  4)( s  2)  3  3s 
2
 s4
 s  1 s  4 s 

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Method 2: Diagonalization
Example 4.5
0   x1  1
 x1    1 0
 x    0  2 0   x   1u
 2 
 2   
0  3  x3  1
 x3   0
 x1 
y  6  6 1 x2 
 
 x3 
e  t

At
Φ (t )  e   0
0

0
e 2 t
0
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diagonal matrix
0 

0 
e 3t 
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Diagonalization via Coordinate Transformation
Plant:
A  R nn
x  Ax  Bu
y  Cx  Du
Eigenvalue of A:
i , satisfying Avi  i vi , i  1,,n
Assume that all the eigenvalues of A are distinct, i.e. 1  2  3  n
Then eigenvectors, v1, v2 ,  ,vn
Coordinate transformation matrix

are independent.
T  [v1, v2 ,  ,vn ]
Λ  T 1 AT is a diagonal matrix.
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0
1



2

T 1 AT    





0

n


where
e At  Te ΛtT 1
e1t

e Λt  


0
e2t
0 




λ nt 
e 
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New coordinate:
  T 1 AT
1  1
 
 2    0
 
 
 n   0


ξ  T -1x
0
0
2
0
0

0

0  1 
0   2 
 
   
n   n 
A  T 1 AT
B  T 1 B
(4.1)
C  CT
i (t )  ii (t ), i  1,,n
Solution of (4.1):
i (t )  e ti (0), i  1, ,n
i
1
x(t )  T (t )  v1e1t ξ1 (0)  v2e2t ξ 2 (0)    vn ent ξ n (0), ξ (0)  T x(0)
The above expansion of x(t) is called modal decomposition.
Hence, system asy. stable ⇔ all the eigenvales of A lie in LHP
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Example
i  distinct
 2 1 
1 
x  
x
,
x
(
0
)


 2
 1  2
 
1  1, 2  3
Find eigenvector
 1  v11 
 1  2
(1I  A)v1  
0



 1  2 v12 
 1
v11  1
   
v12  1
 1  v21 
 3  2
(2 I  A)v2  
0



 3  2 v22 
 1
 v1   1 
   
v2   1
T  v1
1 1 
v2   

1

1


T
1
1  1  1

 2  1 1 
 T 1 AT   1 0 
 0  3


ξ  T 1 x
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 1 0 
ξ  
ξ

 0  3

1  11
2  3 2
(4.2)
1 (0) 
 (0)  T x(0),  (0)  


(
0
)
 2 
1
Solution of (4.1):
x(t )  Tξ (t )  v1e1t ξ1 (0)  v2e2t ξ2 (0)
 3 
 2 
1
ξ (0)  T x (0)  ξ (0)   
1
 
 2
 3  t 1  3t 
2 e  2 e 
3 t 1  1  3t  1 
 x(t )  e      e     3

1

t

3
t

1
2 1  2 
 
 e  e 
2
2

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In the case of A matrix is phase-variable form and
1  2  3  n
P  v1 v2
1
1 
 1





2
n 
 vn    1
 

 n 1
n 1
n 1 
2
n 
1
1

2
1

P AP   



t
e  Pe P
At
3
Vandermonde
matrix
for phase-variable
form





4 
1
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Example: i  distinct
1 0  1
A  0 1 0 
0 0 2 
Repeated eigenvalue case, i.e. i is not distinct
I  A 
 1
0
1
0
 1
0
0
0
 2
 (  1)(  1)(  2)
1  2
0 0 1   v1 
(1 I  A)V1  0 0 0  v2   0
0 0  1 v3 
 v1  1
0v1  0v2  v3  0  v2   0
v3  0
depend
 v1  0
0v1  0v2  v3  0  v2   1
v3  0
V1  V2
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3  2
1 0 1  v1 




(3 I  A)V3  0 1 0 v2   0
0 0 0 v3 
 v1   1
v1  0v2  v3  0  v2    0 
v3   1 
P  V1 V2
1 0  1
1 0 0
V3   0 1 0   P 1 AP  0 1 0
0 0 1 
0 0 2
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Case 3: i  distinct
Jordan form
1  2  3
P  v1
v2
v3   P AP  Jordan
1
form
Generalized eigenvectors
(1 I  A)v1  0
(1 I  A)v2  v1
(1 I  A)v3  v2
e

Aˆ t
e 


1t
1 1



1
ˆ
P AP  A  
1 1 

1 
te
1t
e 1t
e 1t 
1t 
te 
e 1t 
t2
2
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Example:
 3 1
A


 1 1
 3
1
1
 1
 (  2) 2
 1  1 v11 
(1I  A)V1  
0



 1 1  v12 
v11   1 
   
v12   1
 1  1 v21 
1
(1I  A)V2  
  



 1 1  v22 
 1
P  V1
v21  1
   
v22  0
 1 1
2 1 
1
ˆ
V2   
 P AP  A  



1
0
0
2




e 2t
e 

Aˆ t
te2t 
At
Aˆ t 1
 e  Pe P
2t 
e 
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Method 3: Cayley-Hamilton Theorem
Theorem: Every square matrix satisfies its char. equation.
Given a square matrix A, A  R nn. Let f(λ) be the char. polynomial of A.
Char. Equation:
f ( )  n  an 1n 1    a1  a0  0
By Caley-Hamilton Theorem
f ( A)  An  an 1 An 1    a1 A  a0 I  0
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22
An  an 1 An 1    a1 A  a0 I  0
An  an 1 An 1    a1 A  a0 I
An 1  an 1 An    a1 A2  a0 A
  an 1 (an 1 An 1    a1 A  a0 I )   a1 A2  a0 A
any
f ( A)  k0 I  k1 A  k 2 A    k n A  
2
n
f ( A)   0 I  1 A   2 A     n 1 A
2

n 1
n 1

k 0
k
A
k
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Example:
100
A
?
1 2
A

0
1


f ( A)  A100   0 I  1 A
let
 1
2
0
 2
 (  1)(  2)  0 , 1  1, 2  2
f (1 )  1
100
  0  11  1100
 0  2  2100
100
2
f (2 )  
  0  12  2
1  2100  1
100
101
1
0
1
2

1
2
 2




100
100
100
f ( A)  A  (2  2 ) 
 (2  1) 




0
1
0
1
1 



 0
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Example:
 3  1
A

2
0


e ?
At
 3 1
 0 , 1  1, 2  2
2 
f (1)  e t   0  11   0  1
 0  2e t  e 2t
f (2)  e 2t   0  12   0  1 2
1  e 2t  e t
t
e  2e  e
At
 2t
 1
1 0
 2t
t   3
0 1   (  e  e )  2

0




 2e  2 t  e  t

 2t
t

2
e

2
e

e  2 t  e t 
 2t
t 
 e  2e 
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Example:
 2 0  1  1 
x  Ax  bu  4 1  4 x  1u
2 0  1  0
y  1 0 1x
  0,1,1
1 
1   0 
v1  4, v 2  0, 1 
2
1  0
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Example:
 0 1 0  0 
x  Ax  bu  0 0 1  x  0u
2 1  2 1
y  1 0 0x
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